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Group 2Volumetric Flowrate and
Average Velocity
Sage Ross
Randy Goll
Lisa Gourley
Marci Wyatt
Jennifer Oberlag
April 21, 2001
Problem Statement
Fluid flows through a conduit with a quarter circle cross-section. The profile of the velocity in the z-direction can beapproximated by:
])/4(1[ ])/(1[ )/( 22 RrRrvv Az
With the coordinate system as shown and where vA is a constant equal to 20 cm/s and R=30 cm. Find the Volumetric flowrate Q and the average velocity.
The circular shape of the conduit suggests that we use cylindricalcoordinates to solve the problem.
vA=20 cm/s R=30 cm
3 dimensional view of conduit
Quarter circle cross section
])/4(1[ ])/(1[ )/( 22 RrRrvv Az
We need to use the following equations to solve the problem:
A
nvQ dA )(
A
Qvave
A
dAnvQ )(
Solution
zvnv drrddA
R
zvQ0
4
4
dr dr
R
Rr
Rr
A rvQ0
4
43
162 dr - ]1[ 2
3
Volumetric flowrate across element dA: dAnv )(
To find the volumetric flow we must integrate:
Such that: and
The limits for integration can be found from geometry.
Substitute the given value for vz:
R
AvQ0
4
4
242
Rr
Rr dr d]r -][1-[1
Integrate with respect to
1 0
2
3
16
3
16
44 2
64
3
2
64
3
R
Rr
Rr
A rdrvQ
R R
RrrA
RrA
R
vdrr
R
vQ
0 0532
2
53
2
4
312122
22
53 cm 30cm/s 2045
2
15
2
332
53
Rv
R
vQ AR
RRA
sQ /cm 2500 3
Evaluate the integral at the limits:
Simplify and integrate with respect to r:
Simplify and evaluate:
2
3
24
1 cm) 30(
)/cm 2500(4
s
R
Q
A
Qvave
cm/s 5.3avev
The average velocity is found by diving volumetric flowrate by area.
vz
Vave represents the average of all of the velocities over the velocity profile.